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I'm asked to show that $displaystylesum_{r=1}^infty dfrac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$, and am given the hint that for any $z in mathbb{C}backslashmathbb{N}$ the series is uniform on some neighbourhood of $z$.

Someone has asked the same question here:

Showing $sum_{r=1}^infty frac{1}{(r-z)^2}$ is holomorphic on $mathbb{C}backslashmathbb{N}$

but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?

I know I can use Morera's theorem to show that $displaystyleint_{gamma} sum_{r=1}^infty dfrac{1}{(r-z)^2} text{dz} = 0 $ for any closed path $gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.

Thanks in advance.

You can use the following simpler fact. If

• 
  $f_n$ are holomorphic in $Omega$,


• 
  $f_nto f$ uniformly in $Omega$,


• 
  $f_n'to g$ uniformly in $Omega$,

then $f$ is holomorphic with $f'=g$.

Then check that with $f_n$ equal to the partial sum and $OmegaSubsetmathbb Csetminusmathbb N$ the assumptions are verified.